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This question already has an answer here:

I came across this problem while solving another one. I will show how far I could get on my own:

Suppose that $\sqrt 3 \in Q(\sqrt[4]2)$. Since $Q(\sqrt3)$ is the smallest field containing both $Q$ and $\sqrt3$, thus $Q(\sqrt3)\subset Q(\sqrt[4]2)$.

Now $[Q(\sqrt[4]2):Q] =4 $ and $[Q(\sqrt3):Q]=2$ (I've already proven both), so$[Q(\sqrt[4]2):Q(\sqrt 3)] =2$. Therefore by the definition of degree extension, there exist $p(x)\in Q(\sqrt 3)[X]$ of degree 2 realizing $\sqrt[4]2$, that is, exists: $a,b,c\in Q(\sqrt 3)$ such that:

$$a(\sqrt[4]2)^2+b \sqrt[4]2+c =0.$$

How to proceed now? I should use the fact that those extensions lie in $\mathbb R$ to get a contradiction? I only know that $a\neq 0$, but what about the other coefficients?

EDIT: I could find an argument below, check the answers. Check it out if you agree! The argument is just a continuation of the above one.

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marked as duplicate by Watson, Community Nov 23 '18 at 21:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Possibly related: math.stackexchange.com/questions/2995190. A computation-free proof would be : if $\sqrt 3 \in \Bbb Q(\sqrt[4]{p})$ for some prime $p \neq 3$, then we have an inclusion $\Bbb Q(\sqrt p, \sqrt 3) \subset \Bbb Q(\sqrt[4]{p})$, which is an equality because of the degrees (one only needs a small computation with the traces to show that the former field has indeed degree $4$ over $\Bbb Q$ ; the latter has degree $4$ by Eisenstein). […] $\endgroup$ – Watson Nov 23 '18 at 15:56
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    $\begingroup$ […] Then $\Bbb Q(\sqrt[4]{p})$ would be Galois over $\Bbb Q$, since $\Bbb Q(\sqrt p, \sqrt 3) / \Bbb Q$ is. But the conjugate $i \sqrt[4]{p}$ of $\sqrt[4]{p}$ does not belong to $\Bbb Q(\sqrt[4]{p}) \subset \Bbb R$. $\endgroup$ – Watson Nov 23 '18 at 15:56
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    $\begingroup$ (My proof should also show that $\sqrt n \not \in \Bbb Q(\sqrt[4]{m})$ for every coprime integers $n,m>1$). $\endgroup$ – Watson Nov 23 '18 at 16:06
  • $\begingroup$ @Watson That helped a lot! I used the method in that question to solve it. $\endgroup$ – math.h Nov 23 '18 at 21:22
  • $\begingroup$ I asked a similar question before math.stackexchange.com/questions/2431221/…. This might be helpful. $\endgroup$ – Seewoo Lee Nov 24 '18 at 3:23
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I think that I could find an answer. Check it out if it's correct:

As $p(x) = ax^2+bx+c\in Q(\sqrt3)[X]$ is a polynomial of smallest degree killing $\sqrt[4]2$ (by the very definition of being algebraic of degree 2 over $Q(\sqrt3)$), we must have $a\neq 0$. Also we can conclude that $c\neq 0$, otherwise $p(x) = xq(x)$, where degree of $q(x) $ $<$ degree of $p(x)$; and therefore $q(\sqrt[4]2) = 0,$ which again contradicts the minimality of the degree of $p(x)$.

Now we see that $b\neq 0$. If not, noticing that those field extensions all lie in $\mathbb R$, we find that $a\sqrt 2+c=0$ and hence that $\sqrt2\in Q(\sqrt3), $ which does not happen.

Therefore, $p(x)$ is such that $a,b,c$ are all non-zero. Solving for $\sqrt 2$, we find that $$\sqrt2 = \frac{-b\sqrt[4]2-c}{a} \,.$$

Since $\sqrt 2 \not\in Q(\sqrt 3)$, we must have the denominator of the above equation equal to zero and hence $\sqrt[4]2\in Q(\sqrt 3)$. By the minimality argument of fields, we see that $Q(\sqrt 3) = Q(\sqrt[4]2)$ (noticed that we have already assumed that $Q(\sqrt 3) \subset Q(\sqrt[4]2)$). But again this cannot happen, since the degree of $Q(\sqrt[4]2)$ over $Q$ is $4$ and the degree of $Q(\sqrt 3)$ over $Q$ is $2$.

With all this such contradictions, we conclude that $\sqrt 3 \not \in Q(\sqrt[4]2)$.

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$\sqrt3=p/q\sqrt[4]{2} \text{ with } gcd(p,q)=1\Rightarrow 9=(p^4/q^4) 2\Rightarrow 9q^4=2p^4\Rightarrow 3|p \text{ and } 2|q.$ Suppose $p=3a$ and $q=2b$. Then $9\times 16 b^4=2\times 81 a^4\Rightarrow 8b^4=9a^4\Rightarrow 3|b\Rightarrow 3|q\Rightarrow gcd(p,q)>1.$

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